Thermal decomposition of AgNO$$_3$$ produces two paramagnetic gases. The total number of electrons present in the antibonding molecular orbitals of the gas that has the higher number of unpaired electrons is _______.

Silver nitrate decomposes on heating according to

$$2\,AgNO_3 \;\xrightarrow{\;\Delta\;}\; 2\,Ag \;+\; 2\,NO_2 \;+\; O_2$$

The two gaseous products are $$NO_2$$ and $$O_2$$. Both are paramagnetic, but the number of unpaired electrons differs:

• $$NO_2$$ has one unpaired electron (odd total electron count).
• $$O_2$$ has two unpaired electrons (the well-known dioxygen case).

Hence, the gas with the larger number of unpaired electrons is $$O_2$$.

Next, write the molecular orbital (MO) configuration of $$O_2$$ (total 16 electrons):

$$\sigma_{1s}^{2}\;\sigma_{1s}^{*\,2}\; \sigma_{2s}^{2}\;\sigma_{2s}^{*\,2}\; \sigma_{2p_z}^{2}\; \pi_{2p_x}^{2}\;\pi_{2p_y}^{2}\; \pi_{2p_x}^{*\,1}\;\pi_{2p_y}^{*\,1}$$

Antibonding orbitals are those marked with an asterisk ($$^*$$). Counting the electrons in these orbitals:

$$\sigma_{1s}^{*} : 2,\; \sigma_{2s}^{*} : 2,\; \pi_{2p_x}^{*} : 1,\; \pi_{2p_y}^{*} : 1$$

Total electrons in antibonding MOs

$$= 2 + 2 + 1 + 1 = 6$$

Therefore, the required number is 6.